lim n tends to infinity (1^2n+2^2(n-1)+3^2(n-2)+........+n^21)/(1^3+2^3+.....)

To analyze the limit as \( n \) approaches infinity for the expression \[ \frac{1^2 n + 2^2 (n-1) + 3^2 (n-2) + \ldots + n^2 1}{1^3 + 2^3 + \ldots + n^3}, \] we need to break it down into manageable parts. Let's start by simplifying both the numerator and the denominator separately.

Numerator Breakdown

The numerator can be expressed as a sum of terms where each term is weighted by how many times it appears as \( n \) decreases. Specifically, we can rewrite it as:

• For \( k = 1 \) to \( n \), the term \( k^2(n-k+1) \) contributes to the sum.

This gives us:

\[ \text{Numerator} = \sum_{k=1}^{n} k^2 (n - k + 1). \]

Expanding this, we have:

\[ \sum_{k=1}^{n} k^2 n - \sum_{k=1}^{n} k^3 + \sum_{k=1}^{n} k^2. \]

Using the formulas for the sums, we know:

• \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)
• \(\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2\)

Thus, the numerator can be approximated as:

\[ n \cdot \frac{n(n+1)(2n+1)}{6} - \left(\frac{n(n+1)}{2}\right)^2 + \frac{n(n+1)(2n+1)}{6}. \]

Denominator Analysis

Now, let’s look at the denominator:

\[ \text{Denominator} = \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2. \]

Combining the Results

Now we can substitute our findings back into the limit expression:

\[ \lim_{n \to \infty} \frac{\text{Numerator}}{\text{Denominator}}. \]

As \( n \) becomes very large, we can focus on the leading terms in both the numerator and denominator. The leading term in the numerator is approximately \( \frac{n^4}{6} \) and in the denominator, it is \( \frac{n^4}{4} \).

Final Limit Calculation

Now we can compute the limit:

\[ \lim_{n \to \infty} \frac{\frac{n^4}{6}}{\frac{n^4}{4}} = \lim_{n \to \infty} \frac{4}{6} = \frac{2}{3}. \]

Thus, the limit as \( n \) approaches infinity for the given expression is:

Answer: \( \frac{2}{3} \)